UVa 1331 biggest area smallest triangulation

The meaning of problems: Enter a polygon (not necessarily convex hull), to find a maximum area of ​​the smallest triangle triangulation, the maximum output of the triangle.

Analysis: Optimal triangulation. DP [i] [j] represents the value from point i to point j of optimum enumeration intermediate K point . Transfer equation dp [i] [j] = min (dp [i] [j], max (area (i, j, k), max (dp [i] [k], dp [k] [j]) )) . But not necessarily give the title convex polygon, then for any one sub-polygons, arbitrary if any one of a diagonal line of intersection i, j and a polygon edge, then set D (i, j) to infinity , the boundary is d (i, i + 1) = 0.

Reference: https://www.cnblogs.com/Konjakmoyu/p/4905563.html .

Code:

#include<bits/stdc++.h>
using namespace std;
#define INF 0xfffffff
const int Maxm=50+5;

int m;
int x[Maxm],y[Maxm];
double d[Maxm][Maxm];

double area(int a,int b,int c)
{
    double s=(double)(1.0/2)*(x[a]*y[b]+x[b]*y[c]+x[c]*y[a]-x[a]*y[c]-x[b]*y[a]-x[c]*y[b]);
    if(s<0) return -s;
    return s;
}
///面积法判断三角形是否合法（a-b是否为对角线）
bool check(int a,int b,int c)
{
    int i;
    for(i=1;i<=m;i++)
    {
        if(i==a||i==b||i==c) continue;
        double d=area(a,b,i)+area(a,c,i)+area(b,c,i)-area(a,b,c);
        if(d<0) d=-d;
        if(d<=0.01) return 0;
    }
    return 1;
}

int main()
{
    int n;
    scanf("%d",&n);
    while(n--)
    {
        int i,j,k;
        scanf("%d",&m);
        for(i=1;i<=m;i++) scanf("%d%d",&x[i],&y[i]);
        for(i=m;i>=1;i--)
        {
            d[i][i+1]=0.0;
            for(j=i+2;j<=m;j++)
            {
                d[i][j]=INF;
                for(k=i+1;k<j;k++)
                {
                    if(check(i,j,k)) ///判断i j k三点构成的三角形是否合法（三角形内不包含其他点）
                    d[i][j]=min(d[i][j],max(max(area(i,j,k),d[i][k]),d[k][j]));
                }
            }
        }
        printf("%.1lf\n",d[1][m]);
    }
    return 0;
}

Liu Rujia Code:

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;

const double eps = 1e-10;
int dcmp(double x)
{
    if(fabs(x) < eps) return 0;
    else return x < 0 ? -1 : 1;
}

struct Point
{
    double x, y;
    Point(double x=0, double y=0):x(x),y(y) { }
};

typedef Point Vector;

Vector operator + (const Vector& A, const Vector& B)
{
    return Vector(A.x+B.x, A.y+B.y);
}
Vector operator - (const Point& A, const Point& B)
{
    return Vector(A.x-B.x, A.y-B.y);
}
Vector operator * (const Vector& A, double p)
{
    return Vector(A.x*p, A.y*p);
}

bool operator < (const Point& a, const Point& b)
{
    return a.x < b.x || (a.x == b.x && a.y < b.y);
}

bool operator == (const Point& a, const Point &b)
{
    return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;
}

double Dot(const Vector& A, const Vector& B)
{
    return A.x*B.x + A.y*B.y;
}
double Cross(const Vector& A, const Vector& B)
{
    return A.x*B.y - A.y*B.x;
}
double Length(Vector A)
{
    return sqrt(Dot(A, A));
}

bool SegmentProperIntersection(const Point& a1, const Point& a2, const Point& b1, const Point& b2)
{
    double c1 = Cross(a2-a1,b1-a1), c2 = Cross(a2-a1,b2-a1),
           c3 = Cross(b2-b1,a1-b1), c4=Cross(b2-b1,a2-b1);
    return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4)<0;
}

bool OnSegment(const Point& p, const Point& a1, const Point& a2)
{
    return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0;
}

typedef vector<Point> Polygon;

int isPointInPolygon(const Point& p, const Polygon& poly)
{
    int n = poly.size();
    int wn = 0;
    for(int i = 0; i < n; i++)
    {
        const Point& p1 = poly[i];
        const Point& p2 = poly[(i+1)%n];
        if(p1 == p || p2 == p || OnSegment(p, p1, p2)) return -1; // 在边界上
        int k = dcmp(Cross(p2-p1, p-p1));
        int d1 = dcmp(p1.y - p.y);
        int d2 = dcmp(p2.y - p.y);
        if(k > 0 && d1 <= 0 && d2 > 0) wn++;
        if(k < 0 && d2 <= 0 && d1 > 0) wn--;
    }
    if (wn != 0) return 1; // 内部
    return 0; // 外部
}

const int maxn = 100 + 5;

bool isDiagonal(const Polygon& poly, int a, int b)
{
    int n = poly.size();
    for(int i = 0; i < n; i++)
        if(i != a && i != b && OnSegment(poly[i], poly[a], poly[b])) return false; // 中间不能有其他点
    for(int i = 0; i < n; i++)
        if(SegmentProperIntersection(poly[i], poly[(i+1)%n], poly[a], poly[b])) return false; // 不能和多边形的边规范相交
    Point midp = (poly[a] + poly[b]) * 0.5;
    return (isPointInPolygon(midp, poly) == 1); // 整条线段在多边形内
}

const double INF = 1e9;
double d[maxn][maxn];

double solve(const Polygon& poly)
{
    int n = poly.size();
    for(int i = 0; i < n; i++)
        for(int j = 0; j < n; j++) d[i][j] = -1;
    for(int i = n-2; i >= 0; i--)
        for(int j = i+1; j < n; j++)
        {
            if(i + 1 == j) d[i][j] = 0;
            else if(!(i == 0 && j == n-1) && !isDiagonal(poly, i, j)) d[i][j] = INF;
            else
            {
                d[i][j] = INF;
                for(int k = i+1; k < j; k++)
                {
                    double m = max(d[i][k], d[k][j]);
                    double area = fabs(Cross(poly[j]-poly[i], poly[k]-poly[i])) / 2.0; // triangle i-j-k
                    m = max(m, area);
                    d[i][j] = min(d[i][j], m);
                }
            }
        }
    return d[0][n-1];
}

int main()
{
    int T, n;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d", &n);
        double x, y;
        Polygon poly;
        for(int i = 0; i < n; i++)
        {
            scanf("%lf%lf", &x, &y);
            poly.push_back(Point(x,y));
        }
        printf("%.1lf\n", solve(poly));
    }
    return 0;
}